This standing wave resonance occurs when the frequency of oscillation \(f_{SHM}\) (SHM along a radial line), is an integer multiple of the angular frequency, \(f_{cir}\). The results of resonance are sharp peaks at discrete locations along \(x\).
When we flip this graph vertically,
which is the absorption spectrum below a threshold intensity.
\(f_{cir,\,1}=\cfrac{v_{min,\,1}}{2\pi x_1}\)
\(f_{cir,\,2}=\cfrac{v_{min,\,2}}{2\pi x_2}\)
But,
\(\cfrac{f_{SHM}}{f_{cir,\,1}}=N\)
\(\cfrac{f_{SHM}}{f_{cir,\,2}}=M\)
which leads to the proportion identity,
\(\cfrac{x_1}{x_2}\cfrac{v_{min,\,2}}{v_{min,\,1}}=\cfrac{N}{M}\)
For three dip points on the graph, \(X_1,\,\,Y_1\), \(X_2,\,\,Y_2\) and \(X_3,\,\,Y_3\),
\(\cfrac{X_1}{X_2}\sqrt{\cfrac{Y_2}{Y_1}}=\cfrac{N}{M}\)
\(\cfrac{X_1}{X_3}\sqrt{\cfrac{Y_3}{Y_1}}=\cfrac{N}{O}\)
\(\cfrac{X_3}{X_2}\sqrt{\cfrac{Y_2}{Y_3}}=\cfrac{O}{M}\)
From which \(f_{SHM}\) can be calculated, once \(M\), \(N\) or \(O\) has been determined graphically.
You are not going to get a graph of \(v^2_{min}\) versus \(x\), of course.
But in my dream...
